The correct match of each item to the clean water regulation it describes is as follows:
- Regulates pollutants discharged into surface waters: Clean water act
- Covers both surface and ground waters: Safe drinking water act
- Authorizes the EPA to establish minimum standards for tap water: Safe drinking water act
- Funds sewage treatment plants: Clean water act
<h3>What are the functions of clean water regulation?</h3>
Clean Water Act (CWA) is a regulatory body that establishes the basic structure for the regulation of pollutants discharge and maintenance of quality standards of the surface waters.
On the other hand, the Safe Drinking Water Act was founded to oversee the protection of the quality drinking water. The regulatory body is primarily concerned with potable water all waters, whether from above ground or underground sources.
Therefore, the correct match of each item to the clean water regulation it describes is as follows:
- Regulates pollutants discharged into surface waters: Clean water act
- Covers both surface and ground waters: Safe drinking water act
- Authorizes the EPA to establish minimum standards for tap water: Safe drinking water act
- Funds sewage treatment plants: Clean water act
Learn more about clean water regulation at: brainly.com/question/2142268
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Vo= 331+0.6T
360=331+0.6T
360-331=0.6T
29=0.6T
0.6T/29
T=6/290 so change it to simplest form and us formulas good luck
Answer:
Net force exerted on the radio is 27.5 Newton.
Given:
Mass = 5.5 kg
Acceleration = 5 
To find:
Force exerted on the radio = ?
Formula used:
F = ma
Where F = net force
m = mass
a = acceleration
Solution:
According to Newton's second law of motion,
F = ma
Where F = net force
m = mass
a = acceleration
F = 5.5 × 5
F = 27.5 Newton
Hence, Net force exerted on the radio is 27.5 Newton.
Answer:
There is no displacement.
Explanation:
Because the runner is running laps and returning to the original place, there is no displacement as displacement is relative to the change in location from the original position.
Hope this helps. . .
ly UwU
The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:
(1)
The work done is equal to the elastic energy stored by the compressed spring:
where
is the spring constant and
is the compression of the spring. If we substitute the numbers, we find:
And now we can use eq.(1) to calculate the average power output:
